Question

write an equation of each horizontal tangent line to the curve 2y^3+6(x^2)y-12x^2+6y=1.

Answer 1

y=2?

Answer 2

Start by differentiating the equation implicitly: 6y^2 dy/dx + 6x^2 dy/dx + 12xy - 24x + 6 dy/dx = 0. Now solve for dy/dx to get (4x - 2xy)/(x^2+y^2+1). Since you want a horizontal tangent, set dy/dx = 0. This yields 2x(2-y) = 0 so that x = 0 or y = 2. Now plug in these values into the equation of the curve to find the possible locations of the horizontal tangents. When y = 2, we get 28=0 which means that there is no point on the curve when y=2. When x=0, use your calculator to solve 2y^3+6y-1=0. So, at the point (0, 0.165) there is a horizontal tangent. The equation of this line is y=0.165.